A die is thrown twice. What is the probability that

(i) 5 will not come up either time?

(ii) 5 will come up at least once?

Total number of outcomes = 6 × 6

= 36

(i)Total number of outcomes when 5 comes up on either time are (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5)

Hence, total number of favourable cases = 11

P (5 will come up either time)

P (5 will not come up either time)

(ii)Total number of cases, when 5 can come at least once = 11

P (5 will come at least once)

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as the die is rolled twice

6*6=total number of outcomes

=36

now favourable outcome-(5,1) (5,2) (5,3) (5,4) (5,5 )(5,6) and

(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)

=11 outcomes.

so p(E)=favourable outcome /total outcome

p(E)=11/36

p(not E)=1-11/36

=36-11/36=25/36

favourable outcomes=11

total outcome=36

p(E)=11/36

6*6=total number of outcomes

=36

now favourable outcome-(5,1) (5,2) (5,3) (5,4) (5,5 )(5,6) and

(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)

=11 outcomes.

so p(E)=favourable outcome /total outcome

p(E)=11/36

p(not E)=1-11/36

=36-11/36=25/36

**probability of 5 not coming = 25/36**favourable outcomes=11

total outcome=36

p(E)=11/36

**thus probability of 5 coming once=25/36**- 16

as the die is rolled twice

6*6=total number of outcomes

=36

now favourable outcome-(5,1) (5,2) (5,3) (5,4) (5,5 )(5,6) and

(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)

=11 outcomes.

so p(E)=favourable outcome /total outcome

p(E)=11/36

p(not E)=1-11/36

=36-11/36=25/36

probability of 5 not coming = 25/36

favourable outcomes=11

total outcome=36

p(E)=11/36

thus probability of 5 coming once=25/36

?

6*6=total number of outcomes

=36

now favourable outcome-(5,1) (5,2) (5,3) (5,4) (5,5 )(5,6) and

(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)

=11 outcomes.

so p(E)=favourable outcome /total outcome

p(E)=11/36

p(not E)=1-11/36

=36-11/36=25/36

probability of 5 not coming = 25/36

favourable outcomes=11

total outcome=36

p(E)=11/36

thus probability of 5 coming once=25/36

?

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